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"Sample Notebook - Heat and Mass Transfer by R.K. Rajput : Chapter 1 - Basic Concepts\n",
"author: Umang Agarwal\n",
"\n",
"\n",
"# Example 1.1 Page 16-17\n",
"\n",
"L=.045; \t\t \t\t\t#[m] - Thickness of conducting wall\n",
"delT = 350 - 50; \t\t #[C] - Temperature Difference across the Wall\n",
"k=370; \t\t\t\t\t#[W/m.C] - Thermal Conductivity of Wall Material\n",
"#calculations\n",
"#Using Fourier's Law eq 1.1\n",
"q = k*delT/(L*10**6); \t\t\t#[MW/m^2] - Heat Flux\n",
"#results\n",
"print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer per unit area =\",q,\" W\");\n",
"#END\n",
"\n",
"# Example 1.2 Page 17\n",
"\n",
"L = .15; \t\t \t\t\t#[m] - Thickness of conducting wall\n",
"delT = 150 - 45; \t\t #[C] - Temperature Difference across the Wall\n",
"A = 4.5; #[m^2] - Wall Area\n",
"k=9.35; \t\t\t\t\t#[W/m.C] - Thermal Conductivity of Wall Material\n",
"#calculations\n",
"#Using Fourier's Law eq 1.1\n",
"Q = k*A*delT/L; \t\t\t#[W] - Heat Transfer\n",
"#Temperature gradient using Fourier's Law\n",
"TG = - Q/(k*A); #[C/m] - Temperature Gradient\n",
"#results\n",
"print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer per unit area =\",Q,\" W\");\n",
"print '%s %.2f %s' %(\"\\n \\n The Temperature Gradient in the flow direction =\",TG,\" C/m\");\n",
"#END\n",
"\n",
"# Example 1.3 Page 17-18\n",
"\n",
"x = .0825; \t\t \t\t\t#[m] - Thickness of side wall of the conducting oven\n",
"delT = 175 - 75; \t\t #[C] - Temperature Difference across the Wall\n",
"k=0.044; \t\t\t\t\t#[W/m.C] - Thermal Conductivity of Wall Insulation\n",
"Q = 40.5; #[W] - Energy dissipitated by the electric coil withn the oven \n",
"#calculations\n",
"#Using Fourier's Law eq 1.1\n",
"A = (Q*x)/(k*delT); \t\t#[m^2] - Area of wall\n",
"#results\n",
"print '%s %.2f %s' %(\"\\n \\n Area of the wall =\",A,\" m^2\");\n",
"#END\n",
"\n",
"# Example 1.4 Page 18-19\n",
"\n",
"delT = 300-20; \t\t #[C] - Temperature Difference across the Wall\n",
"h = 20; \t\t\t\t\t#[W/m^2.C] - Convective Heat Transfer Coefficient\n",
"A = 1*1.5; #[m^2] - Wall Area\n",
"#calculations\n",
"#Using Newton's Law of cooling eq 1.6\n",
"Q = h*A*delT; \t\t\t#[W] - Heat Transfer\n",
"#results\n",
"print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer =\",Q,\" W\");\n",
"#END\n",
"\n",
"# Example 1.5 Page 19\n",
"\n",
"L=.15; \t\t \t\t\t#[m] - Length of conducting wire\n",
"d = 0.0015; #[m] - Diameter of conducting wire\n",
"A = 22*d*L/7; #[m^2] - Surface Area exposed to Convection\n",
"delT = 120 - 100; \t\t #[C] - Temperature Difference across the Wire\n",
"h = 4500; \t\t\t\t\t#[W/m^2.C] - Convective Heat Transfer Coefficient\n",
"print 'Electric Power to be supplied = Convective Heat loss';\n",
"#calculations\n",
"#Using Newton's Law of cooling eq 1.6\n",
"Q = h*A*delT; \t\t\t#[W] - Heat Transfer\n",
"Q = round(Q,1);\n",
"#results\n",
"print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer =\",Q,\" W\");\n",
"#END\n",
"\n",
"# Example 1.6 Page 20-21\n",
"\n",
"T1 = 300 + 273; \t\t #[K] - Temperature of 1st surface\n",
"T2 = 40 + 273; #[K] - Temperature of 2nd surface\n",
"A = 1.5; #[m^2] - Surface Area\n",
"F = 0.52; \t\t\t\t #[dimensionless] - The value of Factor due geometric location and emissivity\n",
"sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant\n",
"#calculations\n",
"#Using Stephen-Boltzmann Law eq 1.9\n",
"Q = F*sigma*A*(T1**4 - T2**4) \t #[W] - Heat Transfer\n",
"#Equivalent Thermal Resistance using eq 1.10\n",
"Rth = (T1-T2)/Q; #[C/W] - Equivalent Thermal Resistance\n",
"#Equivalent convectoin coefficient using h*A*(T1-T2) = Q\n",
"h = Q/(A*(T1-T2)); #[W/(m^2*C)] - Equivalent Convection Coefficient\n",
"#results\n",
"print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer =\",Q,\" W\");\n",
"print '%s %.2f %s' %(\"\\n The equivalent thermal resistance =\",Rth,\" C/W\");\n",
"print '%s %.2f %s' %(\"\\n The equivalent convection coefficient =\",h,\" W/(m^2 * C)\");\n",
"#END\n",
"\n",
"# Example 1.7 Page 21-22\n",
"\n",
"L = 0.025; #[m] - Thickness of plate\n",
"A = 0.6*0.9; #[m^2] - Area of plate \n",
"Ts = 310; \t\t #[C] - Surface Temperature of plate\n",
"Tf = 15; #[C] - Temperature of fluid(air)\n",
"h = 22; \t\t\t\t\t #[W/m^2.C] - Convective Heat Transfer Coefficient\n",
"Qr = 250; \t\t\t\t #[W] - Heat lost from the plate due to radiation\n",
"k = 45; \t\t\t\t\t #[W/m.C] - Thermal Conductivity of Plate\n",
"#calculations\n",
"# In this problem, heat conducted by the plate is removed by a combination of convection and radiation\n",
"# Heat conducted through the plate = Convection Heat losses + Radiation Losses\n",
"# If Ti is the internal plate temperature, then heat conducted = k*A*(Ts-Ti)/L\n",
"Qc = h*A*(Ts-Tf); #[W] - Convection Heat Loss\n",
"Ti = Ts + L*(Qc + Qr)/(A*k); \t #[C] - Inside plate Temperature\n",
"#results\n",
"print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer =\",Ti,\" C\");\n",
"#END\n",
"\n",
"# Example 1.8 Page 22\n",
"\n",
"Ts = 250; \t\t #[C] - Surface Temperature\n",
"Tsurr = 110; #[C] - Temperature of surroundings\n",
"h = 75; \t\t\t\t\t #[W/m^2.C] - Convective Heat Transfer Coefficient\n",
"F = 1; \t\t\t\t #[dimensionless] - The value of Factor due geometric location and emissivity\n",
"sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant\n",
"k = 10; \t\t\t\t\t #[W/m.C] - Thermal Conductivity of Solid\n",
"#calculations\n",
"# Heat conducted through the plate = Convection Heat losses + Radiation Losses\n",
"qr = F*sigma*((Ts+273)**4-(Tsurr+273)**4) #[W/m^2] - #[W] - Heat lost per unit area from the plate due to radiation\n",
"qc = h*(Ts-Tsurr); #[W/m^2] - Convection Heat Loss per unit area\n",
"TG = -(qc+qr)/k; \t #[C/m] - Temperature Gradient\n",
"#results\n",
"print '%s %.2f %s' %(\"\\n \\n The temperature Gradient =\",TG,\" C/m\");\n",
"#END\n"
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